Q140 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
Two coils have a mutual inductance of $0.005\ \text{H}$. The current changes in the first coil according to equation $I = I_0\sin\omega t$, where $I_0 = 10\ \text{A}$ and $\omega = 60\pi\ \text{rad s}^{-1}$. The maximum values of e.m.f. in the second coil in volt will be
- A$2\pi$
- Correct answerB$3\pi$
- C$4\pi$
- D$6\pi$
Solution
The induced e.m.f. in the second coil is given by $e = -M \dfrac{dI}{dt}$.
Given $I = I_0\sin\omega t$, differentiating with respect to time gives $\dfrac{dI}{dt} = I_0\omega\cos\omega t$.
Thus, the induced e.m.f. is $e = -MI_0\omega\cos\omega t$.
The maximum value of e.m.f. is $e_{\text{max}} = MI_0\omega$.
Substituting the given values $M = 0.005\ \text{H}$, $I_0 = 10\ \text{A}$, and $\omega = 60\pi\ \text{rad s}^{-1}$:
$e_{\text{max}} = 0.005 \times 10 \times 60\pi = 3\pi\ \text{V}$.
Answer: $3\pi$
Given $I = I_0\sin\omega t$, differentiating with respect to time gives $\dfrac{dI}{dt} = I_0\omega\cos\omega t$.
Thus, the induced e.m.f. is $e = -MI_0\omega\cos\omega t$.
The maximum value of e.m.f. is $e_{\text{max}} = MI_0\omega$.
Substituting the given values $M = 0.005\ \text{H}$, $I_0 = 10\ \text{A}$, and $\omega = 60\pi\ \text{rad s}^{-1}$:
$e_{\text{max}} = 0.005 \times 10 \times 60\pi = 3\pi\ \text{V}$.
Answer: $3\pi$
Hence, the correct answer is option: 'B'.